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I presume that answers to the following questions are likely to exist in the literature; so this question is mostly a reference request (but failing that, I would be certainly interested in learning a proof which cannot be readily found in published expositions).

Let $R$ be a Noetherian commutative ring (of finite Krull dimension, if it helps), and let $S=R[x_1,\dotsc,x_n]$ be the ring of polynomials in $n$ variables with the coefficients in $R$. Hilbert's syzygy theorem claims that if $R$ is a field then the global dimension of $S$ does not exceed (in fact, is equal to) $n$. I am interested in the following relative versions of this classical theorem.

Let $G$ be an $S$-module. Assume that $G$ is flat as an $R$-module. How does one prove that the flat dimension of the $S$-module $G$ does not exceed $n$?

Similarly, let $Q$ be an $S$-module. Assume that $Q$ is projective as an $R$-module. How does one prove that the projective dimension of the $S$-module $Q$ does not exceed $n$?

Most interestingly for me, let $K$ be an $S$-module. Assume that $K$ is injective as an $R$-module. How does one prove that the injective dimension of the $S$-module $K$ does not exceed $n$?

[The context: among other things, I am trying to understand the following more general question, which is relevant for my studies of semi-infinite algebraic geometry. Let $R\longrightarrow S$ be a smooth morphism of Noetherian commutative rings. Let $J^\bullet$ be an unbounded complex of injective $S$-modules. Since $S$ is a flat $R$-module, all injective $S$-modules are also injective as $R$-modules. Assume that, as a complex of $R$-modules, $J^\bullet$ is contractible. I would like to show that $J^\bullet$ is then also contractible as a complex of $S$-modules. Notice that this is certainly not true for an arbitrary flat morphism $R\longrightarrow S$, not even for a flat morphism of finite type (take $R$ to be a field).]

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The answers to the first two questions readily follow from Proposition 7.5.2 in [J. C. McConnell and J. C. Robson, "Noncommutative Noetherian rings", AMS, 1987]. Actually, they prove a much more general result which holds for skew polynomial rings and skew Laurent polynomial rings. The base ring $R$ can be arbitrary, not necessarily commutative.

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  • $\begingroup$ Thanks! This is a most relevant reference. $\endgroup$ – Leonid Positselski May 25 at 16:08
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I think that I can answer the first one of the three questions (which I guess is the simplest one), about the flat dimension. The idea is to reduce the problem to the case when $R$ is a field, using the classical Hilbert's syzygy theorem as a black box. The argument proceeds by Noetherian induction in the ring $R$. So we assume that the assertion is true for all the quotient rings of $R$ by nonzero ideals, and prove that it is also true for $R$.

Consider two cases separately.

I. $R$ has zero-divisors. So there are two nonzero elements $a$ and $b$ in $R$ such that $ab=0$.

We have an $R$-flat $S$-module $G$. Then $G/aG$ is an $R/aR$-flat $S/aS$-module and $G/bG$ is an $R/bR$-flat $S/bS$-module. By the assumption of Noetherian induction applied to the rings $R/aR$ and $R/bR$, the flat dimensions of the $S/aS$-module $G/aG$ and the $S/bS$-module $G/bG$ do not exceed $n$.

We have to show that $\operatorname{Tor}^S_{n+1}(M,G)=0$ for all $S$-modules $M$. Using the short exact sequence $0\to aM \to M \to M/aM \to 0$, the problem is reduced to the particular cases when either $aM=0$ or $bM=0$. Assume $aM=0$. Tensoring a flat resolution of the $S$-module $G$ with $R/aR$ over $R$, we obtain a flat resolution of the $S/aS$-module $G/aG$. Hence $\operatorname{Tor}^S_i(M,G)=\operatorname{Tor}^{S/aS}_i(M,G/aG)=0$ for all $i>n$, as desired.

II. $R$ is an integral domain. Denote by $Q$ the field of fractions of $R$.

Let $M$ be an $S$-module. We have $\operatorname{Tor}^S_i(M,G)\otimes_RQ\simeq\operatorname{Tor}^{S\otimes_RQ}_i(M\otimes_RQ,\>G\otimes_RQ)$ for all $i\ge0$. Since $S\otimes_RQ=Q[x_1,\dotsc,x_n]$ is the ring of polynomials in $n$ variables over a field, by the classical Hilbert's syzygy theorem it follows that $\operatorname{Tor}^S_i(M,G)\otimes_RQ$ for all $i>n$. Thus $\operatorname{Tor}^S_i(M,G)$ are torsion $R$-modules for $i>n$.

Let $a\in R$ be a nonzero element. In order to prove that $\operatorname{Tor}^S_{n+1}(M,G)=0$, it suffices to show that (for every $a$) there are no nonzero elements annihilated by $a$ in $\operatorname{Tor}^S_{n+1}(M,G)=0$.

Surely we can assume that $n>0$ (otherwise $R=S$, and there is nothing to prove). Let $0\to\Omega M\to P\to M\to0$ be a short exact sequence of $S$-modules with a projective (or flat) $S$-module $P$ (so $\Omega M$ is "the syzygy module" of $M$). Then we have $\operatorname{Tor}^S_{n+1}(M,G)=\operatorname{Tor}^S_n(\Omega M,G)$.

Both $\Omega M$ and $G$ are $R$-torsionfree modules (and so is the ring $S$). In particular, they are $a$-torsionfree. Consider the left derived tensor products $\Omega M\otimes_S^{\mathbb L}G$ and $(\Omega M/a\Omega M)\otimes_{S/aS}^{\mathbb L}G/aG$, viewed as objects of the derived category of $R$-modules. Then we have a distinguished triangle $$ \Omega M\otimes_S^{\mathbb L}G\overset{a}\longrightarrow \Omega M\otimes_S^{\mathbb L}G \longrightarrow (\Omega M/a\Omega M)\otimes_{S/aS}^{\mathbb L}G/aG \longrightarrow \Omega M\otimes_S^{\mathbb L}G [1]. $$

From the related long exact sequence of cohomology modules we see that if there are $a$-torsion elements in $\operatorname{Tor}^S_n(\Omega M,G)$ then $\operatorname{Tor}^{S/aS}_{n+1}(\Omega M/a\Omega M,G/aG)\ne0$. This would contradict the assumption of Noetherian induction applied to the ring $R/aR$. End of proof.

EDIT: I was asked to justify the existence of the distinguished triangle. Here is the explanation.

Lemma. Let $S$ be a commutative ring, $a\in S$ be a nonzero-dividing (regular) element, and $M$ and $N$ be two $S$-modules containing no nonzero elements annihilated by $a$. Then there is a distinguished triangle in the derived category of $S$-modules $$ M\otimes_S^{\mathbb L}N \overset{a}\longrightarrow M\otimes_S^{\mathbb L}N \longrightarrow M/aM\otimes_{S/aS}^{\mathbb L}N/aN \longrightarrow M\otimes_S^{\mathbb L}N [1]. $$

Proof of Lemma. Firstly, there is a distinguished triangle $N\overset a\to N\to N/aN\to N[1]$. Taking the left derived tensor product with $M$ over $S$, we obtain $$ M\otimes_S^{\mathbb L}N \overset{a}\longrightarrow M\otimes_S^{\mathbb L}N \longrightarrow M\otimes_S^{\mathbb L}N/aN \longrightarrow M\otimes_S^{\mathbb L}N [1]. $$ It remains to construct an isomorphism $M\otimes_S^{\mathbb L}N/aN \simeq M/aM\otimes_{S/aS}^{\mathbb L}N/aN$ in the derived category of $S$-modules. For this purpose, choose a flat resolution of the $S$-module $M$. This is a resolution of an $a$-torsionfree module by $a$-torsionfree modules, so it stays exact after applying the functor $L\longmapsto L/aL$.

Applying this functor to this resolution, we obtain a flat resolution of the $S/aS$-module $M/aM$. Tensoring the former resolution with $N/aN$ over $S$ produces the same complex of $S$-modules (in fact, $S/aS$-modules) as tensoring the latter resolution with $N/aN$ over $S/aS$.

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  • $\begingroup$ Why there exists such distinguished triangle? $\endgroup$ – A.G May 21 at 16:40
  • $\begingroup$ Essentially the same question for the case of zero-divisors: why "Tensoring a flat resolution of the $S$-module $G$ with $R/aR$ over $R$, we obtain a flat resolution of the $S/aS$-module $G/aG$"? $\endgroup$ – A.G May 21 at 16:46
  • $\begingroup$ @A.G Answering your second question: a flat resolution of the $S$-module $G$ is a resolution of a flat $R$-module by flat $R$-modules (because any flat $S$-module is a flat $R$-module, since $S$ is a flat $R$-module). Tensoring a flat $R$-module resolution of a flat $R$-module with any $R$-module, we obtain an exact complex of $R$-modules. $\endgroup$ – Leonid Positselski May 21 at 16:56
  • $\begingroup$ @A.G Furthermore, tensoring an $S$-module with $R/aR$ over $R$ is the same thing as tensoring it with $S/aS$ over $S$ (both the functors amount to $L\longmapsto L/aL$). Tensoring a flat $S$-module with any ring $T$ over $S$ produces a flat $T$-module. In particular, tensoring a flat $S$-module with $S/aS$ produces a flat $S/aS$-module. Does this answer your second question? $\endgroup$ – Leonid Positselski May 21 at 16:58
  • $\begingroup$ Thank you, and sorry for making waste your time on such easy question. I din't notice the flatness of G over R. $\endgroup$ – A.G May 21 at 17:01
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I contacted some people privately and was suggested the following reference, which answers both the projective and injective dimension questions. The flat dimension question surely can be dealt with similarly.

The paper is: S.Eilenberg, A.Rosenberg, and D.Zelinsky, "On the dimension of modules and algebras VIII. Dimension of tensor products", Nagoya Math. J. 12 (1957), https://projecteuclid.org/journals/nagoya-mathematical-journal/volume-12/issue-none/On-the-dimension-of-modules-and-algebras-VIII-Dimension-of/nmj/1118799929.full

The advantage of this classical (1957) approach is that it does not require the ring $R$ to be either Noetherian or commutative. Let $R$ be an arbitrary associative ring; denote by $S=R[x_1,\dotsc,x_n]$ the ring of polynomials in $n$ variables with the coefficients in $R$ (it is presumed that the elements of $R$ commute with the variables $x_j$ in $S$). Then

  • the $S$-projective dimension of any $S$-module $B$ does not exceed $n$ plus the $R$-projective dimension of $B$;
  • the $S$-injective dimension of any $S$-module $C$ does not exceed $n$ plus the $R$-injective dimension of $C$.

This follows from the spectral sequence (I) on page 74, in Section 2 of the paper. To apply this spectral sequence, one has to pick a commutative ring $K$ over which $R$ is an algebra (say, $K=\mathbb Z$) and consider the $K$-algebra of polynomials $\Gamma=K[x_1,\dotsc,x_n]$.

Then one has to use the fact that the projective dimension of the diagonal $\Gamma\otimes_K\Gamma$-bimodule $\Gamma$ does not exceed (in fact, is equal to) $n$. This projective dimension is denoted by $\dim \Gamma$ or $\dim_K\Gamma$ in the paper. The reference to the computation of this dimension is Theorem 6 in Section 4 of the paper.

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A Noetherian induction argument for the question about the injective dimension, similar to (but more complicated than) the Noetherian induction argument for the question about the flat dimension spelled out in my previous answer, can be now found in the preprint "Quasi-coherent torsion sheaves, the semiderived category, and the semitensor product", https://arxiv.org/abs/2104.05517 , proof of Proposition 10.2(b).

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